Question

A Gaussian surface in the form of a hemisphere of radius R = 9.86 cm lies in a uniform electric field of magnitude E = 7.16 N/C. The surface encloses no net charge. At the (flat) base of the surface, the field is perpendicular to the surface and directed into the surface. What is the flux through (a) the base and (b) the curved portion of the surface?

Answer 1

Given that :

radius of the hemisphere, r = 9.86 cm = 0.0986 m

electric field, E = 7.16 N/C

The surface encloses have no net charge. Typically, flux entering is negative & flux leaving would be positive.

(a) the flux through the base which will be given as ::

using an equation,   = - E dA                                                        { eq.1 }

where, dA = circular area of the base = r²

= - E ( r²)                                                                                        { eq.2 }

inserting the values in eq.2,

= - (7.16 N/C) (3.14) (0.0986 m)²

= - 0.2185 Nm²/C

= - 2.18 x 10^-1 Nm²/C

(b) the flux through the curved portion of the surface which will be given as :

Assuming, no net charge in the hemisphere, then it means that the flux leaving through the curved surface is numerically equal to the flux entering at the base.

₂ = -

then, ₂ = - (-2.18 x 10^-1 Nm²/C)

₂ = 2.18 x 10^-1 Nm²/C